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Solid State Physics
Lecture notes by Michael Hilke McGill University (v. 10/25/2006)
- Introduction Contents
- The Theory of Everything
- H 2 O - An example
- Binding
- Van der Waals attraction
- Derivation of Van der Waals
- Repulsion
- Crystals
- Ionic crystals
- Quantum mechanics as a bonder
- Hydrogen-like bonding
- Covalent bonding
- Metals
- Binding summary
- Structure
- Scattering
- Scattering theory of everything
- 1D scattering pattern
- Point-like scatterers on a Bravais lattice in 3D
- General case of a Bravais lattice with basis
- Example: the structure factor of a BCC lattice
- Bragg’s law
- Summary of scattering
- Properties of Solids and liquids
- single electron approximation
- Properties of the free electron model
- Periodic potentials
- Kronig-Penney model
- Tight binging approximation
- approximation Combining Bloch’s theorem with the tight binding
- Weak potential approximation
- Localization
- Electronic properties due to periodic potential
- Density of states
- Average velocity
- and electrons Response to an external field and existence of holes
- Bloch oscillations - Semiclassical motion in a magnetic field - Quantization of the cyclotron orbit: Landau levels - Magneto-oscillations
- Phonons: lattice vibrations
- Mono-atomic phonon dispersion in 1D
- Optical branch
- dispersion Experimental determination of the phonon
- Origin of the elastic constant
- Quantum case
- Transport (Boltzmann theory)
- Relaxation time approximation
- Case 1: F~ = −e E~
- Diffusion model of transport (Drude)
- Case 2: Thermal inequilibrium
- Physical quantities
- Semiconductors
- Band Structure
- semiconductors Electron and hole densities in intrinsic (undoped)
- Doped Semiconductors
- Carrier Densities in Doped semiconductor
- Metal-Insulator transition
- In practice
- p-n junction
- One dimensional conductance
- contact More than one channel, the quantum point
- Quantum Hall effect
- superconductivity
INTRODUCTION
What is Solid State Physics? Typically properties related to crystals, i.e., period- icity. What is Condensed Matter Physics? Properties related to solids and liquids including crys- tals. For example: liquids, polymers, carbon nanotubes, rubber,... Historically, SSP was considers as he basis for the un- derstanding of solids since many of the properties were
derived based on a periodic lattice. However it appeared that most of the properties are very similar independent of the presence or not of a lattice. But there are many exceptions, for example, localization due to disorder or the disappearance of the periodicity.
Even in crystals, liquid-like properties can arise, such as a Fermi liquid, which is an interacting electron system. The same is also true the other way around since liquids can form liquid crystals.
Hence, the study of SSP and CMP are strongly interre- lated and can not be separated. These inter-correlations are illustrated below.
Crystal
Supraconductivity
Opto-effects
Magnetism
Crystal
Metals Insulators Semiconductors properties
The Big Bang of Condensed Matter Physics
In comparison to room temperature (300K' 25meV) these energies are huge. Hence ionic Crystals like NaCl are extremely stable, with binding energies of the order of 1eV.
Ionic crystals
Some solids or crystals are mainly held together by the electrostatic potential and they include the alkali-halides like (N aCl −→ N a+Cl−).
d
FIG. 2: A simple ionic crystal such as N aCl
The energy per ion pair is
Energy Nionpair
= −α
e^2 d
C
dn^
= −α
- 4 eV [d/˚A]
C
dn^
with 6 < n ≤ 12 ,
(7) where α is the Madelung constant and can be calcu- lated from the crystal structure. C can be extracted ex- perimentally from the minimum in the potential energy and typically n = 12 is often used to model the effect of the Pauli principle. Hence, from the derivative of the potential we obtain:
d 0 =
12 C
e^2 α
so that
Energy Nionpair
11 αe^2 12 d 0
How good is this model? See table below:
Quantum mechanics as a bonder
Hydrogen-like bonding
Let’s start with one H atom. We fix the proton at r = o then we know form basic quantum mechanics that
the ground state energy is then given by E^0 =-13.6 eV. What happens if we add one proton or H+^ to the system which is R away. The potential energy for the electron is then
U (r) = −
e^2 r
e^2 |r − R|
The lowest eigenfunction with eigenvalue -13.6eV is
ξ(r) =
π
a 0
e−r/a^0 , (11)
where a 0 is the Bohr radius. But now we have two protons. If the protons were infinitely apart then the general solution to potential 10 is a linear superposi- tion ,i.e., ψ(r) = αξ(r) + βξ(r − R) with a degener- ate lowest eigenvalue of E^0 = E^0 =-13.6 eV. When R is not infinite, the two eigenfunctions corresponding to the lowest eigenvalues Eb and Ea can be approximated by ψb = ξ(r) + ξ(r − R) and ψa = ξ(r) − ξ(r − R). See figure 3.
0
0
V(r) Ψa(r)
Ψb(r)
R
r
FIG. 3: The potential for two protons with the bonding and anti-binding wave function of the electron
The average energy (or expectation value of Eb) is
Eb = 〈ψ b∗ Hψb〉/〈ψ∗ b ψb〉 (12)
= E^0 −
A + B
with (13)
A = e^2
drξ^2 (r)/|r − R| (14)
B = e^2
drξ(r)ξ(|r − R|)/r (15)
drξ(r)ξ(|r − R|), (16)
where 〈·〉 ≡
·dr and similarly,
Ea = E^0 −
A − B
Hence, the total energy for state ψb is now
Ebtotal = Eb + e^2 /R (18)
When plugging in the numbers Ebtotal has a minimum at 1.5˚A. See figure 4
R 0
Antibonding Ea
Eb
Eb tot
Ea tot
Bonding
R [distance between protons]
E [eV] -13.6 eV
FIG. 4: The energies as a function of the distance between them for the bonding and anti-binding wave functions of the electron
This figure illustrates why they are called bonding and anti-bonding, since in the bonding case the energy is low- ered when the distance between protons is reduced as long as R > R 0.
Covalent bonding
Covalent bonds are very similar to Hydrogen bonds, only that we have to extend the problem to a linear com- bination of atomic orbitals for every atom. N atoms would lead to N levels, in which the ground state has a bonding wave-function.
Metals
In metals bonding is a combination of the effects dis- cussed above. The idea is to consider a cloud of electrons only weakly bound to the atomic lattice. The total elec- trostatic energy can then be written as
Eel = −
drn(r)
R
e^2 r − R
R>R′
e^2 R − R′^
dr 1 dr 2
e^2 n(r 1 )n(r 2 ) |r 1 − r 2 |
which corresponds to an ionic contribution of the form
Eel = −
αe^2 2 rs
where rs =
4 πn
and α is the Madelung constant. Deriving this requires quite a bit of effort. On top of this one has to add the kinetic energy of the electrons, which is of the form:
Ekin =
9 π 4
3 ℏ^2
10 mr^2 s
And finally we have to add the exchange energy, with is a consequence of the Pauli principle. The expression for this term is given by
Eex = −
9 π 4
4 πrs
Putting all this together we obtain in units of the Bohr radius:
E =
- 35 a 0 rS
- 1 a^20 r S^2
- 5 a 0 rs
eV/atom (23)
This last expressions leads to a minimum at rs/a 0 = 1.6. We can now compare this with experimental values and the result is off by a factor between 2 and 6. What went wrong. Well, we treated the problem on a semiclassi- cal level, without incorporating all the electron-electron interactions in a quantum theory. This is very very diffi- cult, but in the large density case this can be estimated and a better agreement with experiments is obtained.
Binding summary
There are essentially three effects which contribute to the binding of solids:
- Van der Waals (a dipole-dipole like interaction)
- ionic (Coulomb attraction between ions)
- Quantum mechanics (overlap of the wave-function)
In addition we have two effects which prevent the collapse of the solids:
Scattering theory of everything
← crystal Incident beam Outgoing beam
i k ⋅ r ei k^ '⋅^ r e
ϕϕϕϕ
r
dV k k’
O
[ ]
( ) d^3 ( )
q k k
q r r qr
∝ (^) ∫∫∫ ⋅ i ⋅ V
A n e
where n( r) is the distribution of scatterers
The total scattered wave off V is
FIG. 7: Diffraction set-up (picture from G. Frossati)
This formulation can describe any scattering process in terms of the scattering amplitude A(~q). The entire
Physics is hidden in n(r), which contains the informa- tion on the position of scatterers and their individual scattering distribution and probability. In the following we discuss the most important implications.
1D scattering pattern
Let’s suppose we have a 1D crystal along direction ˆy with lattice spacing a and that the incoming wave ei k~in·~r is perpendicular to the crystal along ˆx. We want to cal- culate the scattered amplitude along direction k~out. By defining ~q = k~out − k~in, we can write the scattering am- plitude as
A(~q) =
V
n(~r)ei~q·~r^ dr^3 (24)
If we assume that the crystal is composed of point-like scatterers we can write:
n(x, y, z) =
C
N
N∑ − 1
n=
δ(y − an)δ(x)δ(z), (25)
where C is simply a constant. Hence, by inserting 25 into 24, and defining ~q = (0, q, 0) we have
A(q) =
C
N
N∑ − 1
n=
eiqan^ =
C
N
1 − eiaqN 1 − eiaq^
=N →∞
C when q = 2πm/a 0 when q 6 = 2πm/a = Cδq, G~ (26)
G^ ~ = ˆy 2 πm/a is the reciprocal lattice and m an integer. If we had a screen along ˆy we would see a diffraction pattern along I(q) = |A(q)|^2 , since what is measured experimentally is the intensity. In this simple case the reciprocal lattice is the same as the real lattice, but with lattice spacing 2π/a instead.
Point-like scatterers on a Bravais lattice in 3D
We start again with the general form of A(~q) from 24 and assume that our 3D crystal is formed by point-like scatterers on a Bravais lattice R~. Hence,
n(~r) =
C
N
N∑ − 1
n=
δ^3 (~r − R~n), (27)
and
A(~q) =
C
N
N∑ − 1
n=
ei~q·^ R~n^ =N →∞
C when ~q ∈ G~ 0 when ~q ∈/ G~
= Cδ~q, G~
(28) In this case the Bravais lattice (or Real-Space) is
R^ ~n = n 1 a~ 1 + n 2 a~ 2 + n 3 a~ 3 (29)
and the reciprocal space G~m (or k-space) can be deduced from ei R~n·^ G~m^ = 1, hence
G^ ~m = m 1 b~ 1 + m 2 b~ 2 + m 3 b~ 3 , (30)
where
b^ ~ 1 = 2 π ~a^2 ×^ a~^3 | a~ 1 · a~ 2 × a~ 3 |
, b~ 2 =
2 π ~a 3 × a~ 1 | a~ 1 · a~ 2 × a~ 3 |
and b~ 3 =
2 π ~a 1 × a~ 2 | a~ 1 · a~ 2 × a~ 3 | (31)
General case of a Bravais lattice with basis
The most general form for n(~r), when the atoms sits on the basis { u~j } along the Bravais lattice R~n is
n(~r) =
C
N
M,N∑ − 1
j=1,n=
fj (~r − R~n − u~j ), (32)
where fj is the scattering amplitude for the atoms on site u~j and is typically proportional to the number of electrons centered on u~j. Now
A(~q) =
C
N
M,N∑ − 1
j=1,n=
fj (~r − R~n − u~j )ei~q·~rd^3 r
C
N
M,N∑ − 1
j=1,n=
fj (~r)ei~q·~rd^3 r · ei~q·^ u~j^ · ei~q·^ R~n
with ~r → ~r + R~n + u~j = CS(~q) · δ~q, G~, (33)
where we have defined the structure factor S(~q) as
S(~q) =
∑^ M
j=
fj (~r)ei~q·~r^ d^3 r · ei~q·^ u~j^ (34)
This is the most general form. It is interesting to re- mark that in most case, experiments measure the inten- sity I(~q) = |A(~q)|^2 , rather than the amplitude.
Example: the structure factor of a BCC lattice
The BCC crystal can be viewed as a cubic crystal with lattice a and a basis. Therefore, all lattice sites are de- scribed by
R^ ~n = n 1 axˆ + n 2 aˆy + n 3 azˆ + ~uj , (35)
where ~u 1 = 0(ˆx + ˆy + ˆz) and ~u 2 = (a/2)(ˆx + ˆy + ˆz). The structure factor then becomes:
S(~q) =
j=1, 2
fj (~r)ei~q·~rei~q·^ u~j^ d^3 r. (36)
As a simplification we suppose that f 1 = f 2 = f , then
S(~q) = (1 + e(ia/2)~q·(ˆx+ˆy+ˆz)) f˜ (~q), (37)
where f˜ (~q) is the Fourier transform of f. Hence,
A(~q) = S(~q) · δ~q, G~ = f˜ (~q)
2 δ~q, G~ if q 1 + q 2 + q 3 is even 0 if q 1 + q 2 + q 3 is odd (38)
PROPERTIES OF SOLIDS AND LIQUIDS
The Theory of everything discussed in the first section can serve as a guideline to illustrate which part of the Hamiltonian is important for a given property. For in- stance, when interested in the mechanical properties, the terms containing the electrons can be seen as a pertur- bation. However, when considering thermal conductivity, for example, the kinetic terms of the ions and the elec- trons are important. In the following we will start by considering a few cases and we will start with the single electron approximation.
single electron approximation
The single electron approximation can be used to de- rive the energy and density of the electrons. This sim- plest model will always serve as a reference and in some cases the result is very close to the experimental value. Alkali metals (Li, Na, K,..) are reasonably well described by this model, when we suppose that the outer shell elec- tron/atom (there is 1 for Li, Na, K) is free to move inside the metal. This picture leads to the simple example of N electrons in a box. This box can be viewed as a uniform and positively charge background due to the atomic ions. We suppose that this box with size L×L×L has periodic boundary conditions, i.e.,
H = −
∑^ N
n=
ℏ^2 ∇^2 n 2 me with Ψ(0) = Ψ(L) (44)
For one electron the solutions can be written as
Ψ(~r) = ei ~k·~r with E =
ℏ^2 |~k|^2 2 me
and ~k =
2 π L
(n 1 , n 2 , n 3 ), (45) where ni are the quantum numbers, which are positive or negative integers. Since electrons are Fermions we cannot have two electrons in the same state, except for the spin degeneracy. Hence each electron has to have different quantum numbers. This implies that 2π/L is the minimum difference between two electrons in k-space, which means that 1 electron uses up
( 2 π L
)D
of volume in k-space, if D is the dimension of the space. If we now want to compute the electron density (number of electrons per unit volume ne = N/LD^ ) in the ground state, which have k < kF (Fermi sphere with radius kF ), we obtain:
ne = 2 · V (^) kDF ·
L
2 π
)D
/LD^ , (47)
where the pre-factor 2 comes from the spin degeneracy. Hence,
n^3 eD =
k^3 F 3 π^2
in D = 3 since V (^) kDF =
4 π 3
k F^3 (48)
n^2 eD =
k F^2 2 π in D = 2 since V (^) kDF = πk F^2 (49)
n^1 eD =
2 kF π
in D = 1 since V (^) kDF = 2kF , (50)
where the maximum energy of the electrons is
EF =
ℏ^2 k^2 F 2 me the Fermi energy (51)
This defines the Fermi energy: it is the highest energy when all possible states with energy lower than EF are occupied, which corresponds to the ground state of the system. This is one of the most important defi- nitions in condensed matter physics. The electron density can now be rewritten as a function of the Fermi energy, through eqs. (51) and (48) in 3D:
n(EF ) =
(2mEF )^3 /^2 3 π^2 ℏ^3
We now want to define the energy density of states D(E) as
n(EF ) ≡
∫ EF
0
D(E)dE =⇒ D(E) =
∂E
n(E), (53)
which leads to
D(E) =
m
2 mE ℏ^3 π^2 in 3D (54)
D(E) =
m ℏ^2 π in 2D (55)
D(E) =
2 m ℏ^2 π^2 E
in 1D (56)
and illustrated below.
1D
2D
3D
D(E)
E
FIG. 11: Density of states in 1D, 2D and 3D
Properties of the free electron model
Physical quantities at T=0:
- Average energy per electron: 〈E n 〉= 35 EF
- Pressure: P = − ∂(〈 ∂VE〉 V^ )= 25 nEF , where 〈E〉V is the total energy.
- Compressibility κ−^1 = −V ∂P∂V = 23 nEF
Case 2: T 6 = 0: In equilibrium
fF D =
e(E−μ)/kT^ + 1
where the chemical potential is the energy to add one electron μ = FN +1 − FN. μ = EF at T=0 and F is the free energy. The Sommerfeld expansion is valid for kT << μ and
〈H〉 =
0
dEH(E)FF D (E) (58)
∫ (^) μ
0
H(E)dE +
π^2 6
(kT )^2 H′(μ) + ... (59)
〈E〉 =
0
dEED(E)FF D (E) (60)
∫ EF
0
ED(E)dE + π^2 6
(kT )^2 D(EF ) + ...(61)
〈n〉 = n(T = 0) (62)
⇒ μ(T ) ' EF −
π^2 6
(kT )^2
D′(EF )
D(EF )
Physical quantities at T 6 =0: Specific heat
CV =
V
∂〈E〉
∂T
μ,V
π^2 3
k^2 T D(EF ) (64)
Hence, C TV = γ, which is the Sommerfeld parameter.
Periodic potentials
The periodicity of the underlying lattice has important consequences for many of the properties. We will walk through a few of them by starting with the simplest case in 1D.
Kronig-Penney model
Let us first consider the following simple periodic po- tential in 1D.
H = −
ℏ^2 ∇^2
2 m
− V
n
δ(x − na) (65)
The solutions for na < x < na+a are simply plane waves and can be written as
ψ(x) = Aneikx^ + Bne−ikx. (66)
Now the task is to use the boundary conditions in order to determine An and Bn. We have two conditions:
(1) ψ(na − ≤) = ψ(na + ≤) when ≤ → 0 and (67) (2)
∫ (^) na+≤ na−≤ (H^ −^ E)ψ(x)dx^ = 0 =⇒ −^
ℏ 2 m (ψ
′(na + ≤) − ψ′(na − ≤)) = V ψ(na).
Inserting 66 into condition (1) yields
{ An− 1 eikna^ + Bn− 1 e−ikna^ = Aneikna^ + Bne−ikna^ = ψ(na) = ψn An− 1 eiknae−ika^ + Bn− 1 e−iknaeika^ = ψn− 1
e−ikaψn − ψn− 1 = Bn− 1 e−ikan^
e−ika^ − eika
eikaψn − ψn− 1 = An− 1 eikan^
eika^ − e−ika
and
e−ikaψn+1 − ψn = Bne−ikane−ika^
e−ika^ − eika
eikaψn+1 − ψn = Aneikaneika^
eika^ − e−ika
When taking the derivative of 66 then condition (2) implies
2 m
ik Aneikan ︸ ︷︷ ︸ ψn+1−e−ikaψn 2 i sin(ka)
−ik Bne−ikan ︸ ︷︷ ︸ ψn+1−eikaψn − 2 i sin(ka)
−ik An− 1 eikan ︸ ︷︷ ︸ eikaψn−ψn− 1 2 i sin(ka)
+ik Bn− 1 e−ikan ︸ ︷︷ ︸ e−ikaψn−ψn− 1 − 2 i sin(ka)
= V ψn, (71)
〈φl(r − Rm)|H|φl(r − Rn)〉 = −tl
~i=ˆx,ˆy,zˆ
δm,n+~i + δm,n−~i
(^) + ≤lδm,~~n (81)
Hence, only the nearest neighbor in every direction is taken to be non-zero. Further, we assume that there is no overlap between levels of the one atom potential, which allows us to look for a general solution of the following form for each energy level ≤l.
ψl(r) =
~n
cl~nφl(r − R~n) (82)
with eigenvalue El(k). To calculate El(k) we plug-in this Ansatz into eq. (82) and obtain an equation for the coefficients cl~n, which leads to the following equation when using the tight binding approximation given in eq. (81).
clm~≤l − tl
~i=±x,ˆy,ˆˆz
clm~+~i = El(k)clm~ (83)
The solution of this equation are plane waves, which can be verified readily by taking clm~ = ei ~k· R~m and plug- ging it into the equation to obtain
El(k) = ≤l − tl(2 cos(kxax) + 2 cos(ky ay ) + 2cos(kz az ), (84) where ~a is the lattice constant in all 3 space directions: ~a = Rm~+ˆa − Rm~. The energy diagram is illustrated in fig.. The degeneracy of each original single atomic en- ergy level ≤l is lifted by the coupling to the neighboring atoms and leads to a dispersion curve or electronic band structure.
-1.0 -0.5 0.0 0.5 1.
ε 3
ε 2
−π/a k π/a
E
ε 1
FIG. 13: Dispersion curve for the tight binding model
This tight binding approximation is very successful in describing the electrons which are strongly bound to the atoms. In the opposite limit where the electrons or more plane-wave like, the weak potential approximation is more accurate:
Combining Bloch’s theorem with the tight binding approximation
The tight binding approximation is very general and can be applied to almost any system, including non- periodic ones, where the tight binding elements can be assembled in an infinite matrix. For the periodic case, on the other hand, it is possible to describe the system with a finite matrix in order to obtain the full disper- sion relation. This is obtained by combining the Bloch theorem for periodic potentials, where the wave-function from (79) is again:
ψk(~r) = ei ~k·~r uk(~r)
Instead of writing (82) we write the Bloch-tight- binding solution as
ψkl (r) =
~n
ei ~k·R~n cl~nφl(r − R~n),
which now depends explicitly on the wavevector ~k. Using Bloch’s theorem this implies that
cl~n = clm~,
whenever ~n and m~ are related by a linear combination of Bravais vectors. Moreover, the tight binding equation in (83) is the same but with cl~n replaced by cl~nei ~k·R~n
A
B
a
FIG. 14: Diatomic square crystal
We apply this to the simple example of a diatomic square lattice of lattice constant a with alternating atoms A and B shown in figure. We will further assume that we have only one band l. Hence, the Bloch-tight-binding solution is written as
ψk(r) =
~n
ei ~k·R~n c~nφ(r − R~n),
where c~n takes on only to possible values due to Bloch’s theorem: cA or cB. This leads to the following simplified tight binding equations (assuming 〈φl(r − Rn)|H|φl(r − Rn)〉 = ≤A or ≤B and t = −〈φl(r − Rn)|H|φl(r − Rm)〉 when n and m are nearest neighbors):
cA≤A − t
~i=±ˆx,yˆ cB^ e ia~k·~i (^) = E(~k)cA
cB ≤B − t
~i=±x,ˆyˆ cAe ia~k·~i (^) = E(~k)cB
It is now quite straightforward to rewrite these equa- tions in matrix form: ( ≤A −t · g −t · g∗^ ≤B
cA cB
= E(~k)
cA cB
with g = eia ~k·ˆx
- e−ia ~k·xˆ
- eia ~k·ˆy
- e−ia ~k·ˆy
. The disper- sion relation or band structure is then simply given by obtaining the eigenvalues of HBT B , where
HBT B =
≤A −t · g −t · g∗^ ≤B
Weak potential approximation
In this case we consider the effect of the periodic po- tential V (~r) as a perturbation on the plane wave solution
ψ^0 k(~r) = ei ~k·~r , with corresponding energies ≤^0 k = ℏ
(^2) k 2 2 m and Hamiltonian H 0 , i.e., H 0 ψ^0 k(~r) = ≤^0 kψ k^0 (~r). Since the ori- gin of this energy dispersion relation can be chosen from any site of the reciprocal lattice, we have ≤^0 k+K = ≤^0 k, hence these energies are degenerate. This implies that we have to use a degenerate perturbation theory. The mathematical procedure is very similar to the tight bind- ing approximation, but we now expand the solution ψ(~r) of the full Hamiltonian H = H 0 + V (~r) in terms of a sum of plane waves ψ^0 k(~r). Since ≤^0 k+K = ≤^0 k we will only use two plane waves in this expansion: ψ k^0 (~r) and ψ^0 k+K (~r). Hence,
ψ(~r) = αψ k^0 (~r) + βψ^0 k+K (~r), (85)
where the coefficients α and β have to be determined in order to solve Schr¨odinger’s equation:
(H − E)ψ(~r) = 0. (86)
We can find the solution by first multiplying (86) by (ψ k^0 )∗(~r) and then integrating the equation over the whole space which will lead to one equation, and we obtain a second equation by multiplying (86) by (ψ k^0 +K )∗(~r) and then integrating of the whole space. This leads to
α
d^3 r(ψ^0 k)∗(~r)(H − E)ψ^0 k(~r) + β
d^3 r(ψ^0 k)∗(~r)(H − E)ψ^0 k+K (~r) = 0 α
d^3 r(ψ k^0 +K )∗(~r)(H − E)ψ^0 k(~r) + β
d^3 r(ψ k^0 +K )∗(~r)(H − E)ψ^0 k+K (~r) = 0
where (assuming normalized plane waves in the integrals)
d^3 r(ψ k^0 )∗(~r)(H − E)ψ^0 k(~r) =
∫ d^3 re−ikr^ (H^0 +^ V^ (r)^ −^ E)eikr^ =^ ≤^0 k^ + 0^ −^ E d^3 r(ψ k^0 )∗(~r)(H − E)ψ^0 k+K (~r) =
d^3 re−ikr^ (H 0 + V (r) − E)ei(k+K)r^ =
∫ d^3 reiKr^ (≤^0 k+K^ +^ V^ (r)^ −^ E) =^ VK ∫ d^3 r(ψ^ k^0 +K^ )∗(~r)(H^ −^ E)ψ^0 k+K^ (~r) =^ ≤^0 k+K^ −^ E d^3 r(ψ k^0 +K )∗(~r)(H − E)ψ^0 k(~r) = V−K , (88)
and V (^) K~ =
d^3 rei K~·~r V (~r) (the Fourier transform),
d^3 reiKr^ = 0 (for K 6 = 0), and
d^3 rV (~r) = 0. With coefficients (88), equation (87) leads to the following couple of equations: { α(≤^0 k − E) + βVK = 0 αV−K + β(≤^0 k+K − E) = 0
≤^0 k − E VK V−K ≤^0 k+K − E
∣ = 0^ ⇒^ E^ =^
≤^0 k +≤^0 k+K 2 ±
(≤^0 k −≤^0 k+K )^2 4 +^ VK^ V−K^.^ (89)
Finally if ≤^0 k = ≤^0 k+K , we have E = ≤^0 k ± |VK |, which leads to a splitting 2|VK | of the energy levels at these degenerate energies. For the example in figure , this weak potential approximation would give us a splitting of 2 |VK=2π/a| at k = −π/a and k = K − π/a = π/a. This implies that the first order calculation of the energy split- ting due to the weak periodic potential V (r) is equal to twice the fourier transform of this potential evaluated at
the wavevector which corresponds to the two dispersion curves which led to the degenerate energy level.
Localization
When, instead of having a purely periodic potential disorder is included into the system, we no more have
of an electron is q = −e but if ∂
(^2) E ∂k^2 <^ 0 then q is positive (+e). In this case we describe the particles as holes. They represent missing electrons. With these definitions of q and m∗, which is also called the band mass, the semiclassical equations of motion of single electrons in a periodic potential are simply given by eqs. (95) and (96). In general, the effective mass is given by a tensor de- fined as
m∗ αβ = ℏ^2
∂^2 E
∂kα∂kβ
− 1 , (98)
where α and β are the spatial directions. An important consequence of this semiclassical description of the mo- tion of electrons is the dependence of the effective mass on the energy and the band structure. In some cases the effective mass can even diverge (when ∂
(^2) E ∂k^2 = 0).^ Sim- ilarly the sign of the carriers also depends on the band structure and the energy of the carriers. By definition we call the bottom of an energy band and electronic band when ∂
(^2) E ∂k^2 >^ 0 and a hole band when at the top of the energy band ∂
(^2) E ∂k^2 <^ 0.
Bloch oscillations
In the presence of an electric field and a periodic po- tential we can use the equation of motion (96), i.e.,
ℏ k˙ = −eEel ⇒ k = − eEel ℏ
t, (99)
but in a periodic potential and in the tight binding ap- proximation the energy is given by E(k) = − 2 t 0 cos(ka), where a is the lattice constant and t 0 the nearest neighbor overlap integral. Hence, since v = ˙r and 〈v〉 = ℏ−^1 ∂E/∂k we have
〈r〉 =
2 t 0 eEel
cos(
aeEelt ℏ
This means that the average position of the electrons oscillates in time (Bloch oscillations). In artificial struc- tures these Bloch oscillations are typically of the order of 1THz.
Semiclassical motion in a magnetic field
In the presence of a magnetic field (B), we can describe the semiclassical trajectories in k-space using (96), i.e.,
ℏ k˙ = q〈v〉 × B. (101)
Hence, only the values of k perpendicular to the magnetic field will change, which we denote by k⊥. The component parallel to the field, k‖ is not affected by B. During a small time difference
δt = t 2 − t 1 =
∫ (^) t 2
t 1
dt =
∫ (^) k 2 =k(t 2 )
k 1 =k(t 1 )
dk⊥/| k˙|. (102)
Using (101) and (94) and since k˙ is perpendicular to 〈v〉 and B, ˙(k) ∼ ∂E/∂k‖ we obtain
δt =
ℏ^2
qB
∫ (^) k 2
k 1
dk⊥ ∂E/∂k‖
=
ℏ^2
qB
d dE
∫ (^) k 2
k 1
k‖dk⊥. (103)
For a complete turn this leads to
T =
ℏ^2
qB
d dE
k‖dk⊥ ︸ ︷︷ ︸ S
Here S is the area enclosed by an orbit in k-space. This orbit corresponds to an equipotential line perpendicular to the magnetic field. Let’s suppose for simplicity that the effective mass ten- sor m∗^ is diagonal and given by
m∗^ =
mx 0 0 0 my 0 0 0 mz
and that the energy dispersion is harmonic (which is usu- ally true at a band extremum, i.e.,
E(k) =
ℏ^2 k^2 x 2 mx
ℏ^2 k^2 y 2 my
ℏ^2 k^2 z 2 mz
If we assume that the magnetic field is along z and that the average effective mass perpendicular to B is given by m⊥, we can rewrite (106) as
E(k) =
ℏ^2 k ⊥^2 2 m⊥
ℏ^2 k^2 ‖ 2 m‖
where k ⊥^2 = k x^2 +k^2 y. Using (104) and (107 we then obtain
S = πk^2 ⊥ = π(2m⊥E/ℏ^2 ) − πk ‖^2 m⊥/m‖
⇒
dS dE
π 2 m⊥ ℏ^2 ⇒ ωc =
2 π T
qB m⊥
which is the cyclotron frequency. Hence, the cyclotron frequency depends on the average effective mass perpen- dicular to the magnetic field. This allows us to measure the effective mass along different directions, simply by changing the direction of the magnetic field and by mea- suring the cyclotron frequency.
Quantization of the cyclotron orbit: Landau levels
In quantum mechanics the energies of these cyclotron orbits become quantized. To see this we can write the
Hamiltonian of an electron in a magnetic field in the har- monic approximation (107) as
H =
2 m‖
P ‖^2 +
2 m⊥
(P⊥ + qA)^2 , (109)
where B = ∇ × A ⇒ A = −Byˆx in the Landau Gauge if B is along ˆz. In analogy to the harmonic oscillator, the eigenvalues of (109) are then given by
En,k‖ =
ℏ^2 k^2 ‖ 2 m‖
qB m⊥ ︸︷︷︸ ωc
These eigenvalues can be found by writing the wavefunction as ψ = eikxxφn(y − y 0 )eik‖z with y 0 = − ℏ qBkx , which leads to Hψ = ( ℏ^2 k^2 ‖ 2 m‖ +^
P (^) y^2 2 m⊥ +^
1 2 m⊥
qB m⊥
(y − y 0 )^2
φn(y − y 0 ).
In the y direction this is simply the harmonic oscillator with eigenvalues (n + 1/2)ℏωc and in the direction parallel to the field we have a plane wave so that the total energy is given by (110). The quantized levels (n + 1/2)ℏωc due to the magnetic field are called the Landau levels.
Magneto-oscillations
The quantization of the energy levels in the presence of a magnetic field will lead to oscillations of almost any experimental quantity (resistance, thermal conductivity, magnetization, ...) as a function of B. These oscillations can be understood by looking at the density of states due to the quantization (110). Indeed, (110) leads to a peak in the density of states whenever E = (n + 1/2)ℏωc. Therefore, whenever the component of the Fermi energy perpendicular to the magnetic field is equal to one of these quantized levels there is an extremum in the quan- tity measured. The distance between two of these ex- trema is given by
qB 1 m⊥
(n 1 + 1/2) = E⊥ F = ℏ qB 2 m⊥
(n 2 + 1/2)
B 1
B 2
q m⊥
( E F⊥
2 πm^ ℏ^2 ⊥ ·S
)−^1 (n 2 − n 1 )
B
2 πq ℏ
S−^1 , (111)
where the maximum S is S =
k‖dk⊥ = π(kxF )^2 =
π(kFy )^2 and E F⊥ = ℏ
(^2) (kFx ) 2 2 m⊥ =^
ℏ^2 S 2 πm⊥. Hence, the Magneto- oscillations are periodic in 1/B and the period depends on S. An example of these oscillations in resistance is shown in figure 15 as a function of B and 1/B.
(^0) 0.0 0.5 1.
50
100
150
200
250
300
350
400
450
500
1.1 1.3 1.5 1.8 2.0 2.2 2.4 2.6 2.
0
100
200
300
Rxx [
Ω
]
B [T]
Rxx [
Ω]
1/B [T-1]
FIG. 15: Magneto-resistance oscillations in a GaAs/AlGaAs quantum well.
PHONONS: LATTICE VIBRATIONS
In general:
M u¨l = −
m
φlmum, (112)
where ul are the deviations from the original lattice sites and φlm are the elastic constants who have to obey this sum rule
m φlm^ = 0 (translation invariance). In words, equ. 112 simply means that the force producing the devi- ation on lattice site l only depends on the deviations from the other lattice sites. No deviation=no force (equilib- rium). This equation is very general and does not assume that we have a periodic lattice, but in order to calculate things we will use a periodic lattice and start with 1D. In fig. 16 we illustrate two typical displacement waves in 2D.
u (^) s − 1 u (^) s us (^) + 1 u (^) s + 2^ us^ + 3 s +^4 u K
s − 1 s s + 1 s + 2 s + 3 s + 4
u (^) s − 1 u (^) s us (^) + 1 u (^) s + 2^ us^ + 3 s +^4 u K
s − 1 s s + 1 s + 2 s + 3 s + 4
us (^) − 2^ us^ − 1^ u^ s^ us^ + 1 us + 2
K
us (^) − 2^ us^ − 1^ u^ s^ us^ + 1 us + 2
K
FIG. 16: Two types of lattice displacement waves in 2D, trans- verse and longitudinal modes
ℏωin(~k) − ℏω′ out(k~′) ︸ ︷︷ ︸ in/out particle
= ±ℏΩ( K~)
phonon
and the momentum conservation as
ℏ~k − ℏk~′^ = ±ℏ K~ + ℏ G,~ (120)
where G~ is a reciprocal lattice vector. A neutron scat- tering experiment is illustrated below.
FIG. 19: Inelastic neutron scattering experiment in Chalk River used to determine the phonon dispersion
Origin of the elastic constant
From the TOE we have the potential term due to the interactions between ions.
Hions = −
N∑ions
i
ℏ^2 ∇^2 i 2 mi
N∑ions
i<j
qiqj |ri − rj |
This leads to a potential of the form
φ(r 1 ,... , rn) = φ(r^01 ,... , r^0 n) ︸ ︷︷ ︸ Cohesive energy Equilibrium pos.
i
∂φ ∂ri
r^0 i
displacment ︷︸︸︷ ui
︸ ︷︷ ︸ 0
i,j
∂^2 φ ∂ri∂rj
r^0 i ︸ ︷︷ ︸ φi,j
uiuj + · · · (122)
The linear term has to be zero for stability reasons (no minimum in energy otherwise). The classical equation of motion is then simply given by equ. 112, because F^ ~ = −∇~φ. To solve this equation in general, one can write a solution of the form
u ~l = ~≤ · ei ~k· R~l−iωt , (123)
where R~l are the lattice sites. One then has to solve for the dispersion relation ω(~k) in all directions, as illus-
trated in fig. 18. From 123 we have
mω^2 ~≤ =
m
φl,mei ~k·( R~l− R~m)
︸ ︷︷ ︸ φ^ ˆ(k)
where φˆ(k) is a 3 × 3 matrix, since φlm too. This leads to an eigenvalue equation for ω^2 with three eigenvalues: ω L^2 (for ~k ‖ ~≤) and ω T^2 (1,2) (for ~k ⊥ ~≤). Hence, we have on longitudinal mode and two transverse modes and all phonon modes can be described by a superposition of these. If we have two different masses, we have two addi-
tional equations of the form (124). This leads to the general case, where one has two branches the optical one and the acoustic one and each is divided in longitudinal and transverse modes. The acoustic branch has always a dispersion going to zero at k = 0, whereas for the op- tical mode at k = 0 the energy is non-zero. The acous- tic phonons have a gapless excitation spectrum and can therefore be created at very low energy as opposed to optical phonons which require a miniumum energy to be excited.
Quantum case
While many of the properties, like the dispersion rela- tion, can be explained in classical terms, which are simply vibrations of the crystal ions, others, such as statistical properties need a quantum mechanical treatment. Phonons can simply be seen as harmonic oscillators which carry no spin (or spin 0). They can, therefore, be accurately described by bosons with energy
En(k) = ℏω(k)(n + 1/2). (125)
We can now calculate the density of states of phonons, or the number of states between ω and ω + dω, i.e.,
D(ω)dω =
(2π)^3
∫ (^) ω+dω
ω
dk (126)
but dω = ∇kωdk, hence
D(ω) =
(2π)^3
ω=const
dSω |∇kω|
Let us now suppose that we have 3 low energy modes with linear dispersion, one longitudinal mode ωL(k) = cLk and two transverse modes ωT (k) = cT k, hence ∇ωL,T (k) = cL,T , which implies that
dS = 4πk^2. This leads to a density of state
DL,T (ω) =
k^2 2 π^2 cL,T
ω^2 2 π^2 c^3 L,T
or
Dtot(ω) =
ω^2 2 π^2
c^3 L
c^3 T
1 c^3
in the isotropic case. From statistics we know that the probability to find a state at E = En is given by the Boltzmann distribution Pn ∼ e−En/kB^ T^ , with normalization
Pn = 1. Hence,
Pn = e−nℏω/kB^ T^ (1 − e−ℏω/kB^ T^ ) (130)
simply from the normalization condition. We can now calculate the average energy at ω,
E(ω) =
n
EnPn = (1 − e−ℏω/kB^ T^ )ℏω
∑^ ∞
n=
(n + 1/2)(e−ℏω/kB^ T^ )n
= ℏω
︸^ e ℏω/k︷︷B^ T −^1 ︸ 〈n〉
here 〈n〉 is the expectation value of quantum number n at T , which is nothing else but the Bose Einstein distri- bution. As is well known, Bosons obey the Bose-Einstein statistics, where he Bose-Einstein distribution function is given by
fBE =
eE/kB^ T^ − 1
Important: at T = 0, fBE is simply a delta function δ(E). This is the Bose-Einstein condensation, where the
zero energy state is totally degenerate (this does sup- pose that there are no interactions between bosons, with interactions the delta function will be a little broader). The total average energy can then be calculated from the density of states as:
〈E〉 =
dωD(ω)E(ω) (133)
This also allows us to calculate the specific heat CV =
